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Transcript

Misc 12 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4π‘Ÿ/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let ∠ BOC = ΞΈ Now, AC = AO + OC H = r + r cos ΞΈ H = r (1 + cos ΞΈ) And, R = r sin ΞΈ We need to maximize volume of cone. Volume of the cone is V = 1/3 πœ‹π‘…^2 𝐻 V = 1/3 πœ‹π‘Ÿ^2 sin^2⁑"ΞΈ" r (1 + cos ΞΈ) V = 𝟏/πŸ‘ 𝝅𝒓^πŸ‘ γ€–π’”π’Šπ’γ€—^𝟐⁑"ΞΈ" (1 + cos ΞΈ) Finding 𝒅𝑽/π’…πœ½ 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 (𝑑[sin^2⁑〖θ γ€— (1 + cos⁑θ))/π‘‘πœƒ 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 [2 sin⁑〖"ΞΈ" cos⁑〖"ΞΈ" (1+cos⁑〖"ΞΈ" )+sin^2⁑〖"ΞΈ" (βˆ’sin⁑〖"ΞΈ" )γ€— γ€— γ€— γ€— γ€— ] 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 (2 sin⁑〖"ΞΈ" cos⁑〖"ΞΈ" (1+cos⁑〖"ΞΈ" )βˆ’sin^3⁑"ΞΈ" γ€— γ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 (2 sin⁑〖"ΞΈ" cos⁑〖"ΞΈ" (1+cos⁑〖"ΞΈ" )βˆ’sin^3⁑"ΞΈ" γ€— γ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (2 cos⁑〖"ΞΈ" +γ€–2 cos^2〗⁑〖"ΞΈ" βˆ’γ€–π’”π’Šπ’γ€—^𝟐⁑"ΞΈ" γ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (2 cos⁑〖"ΞΈ" +γ€–2 cos^2〗⁑〖"ΞΈ" βˆ’(πŸβˆ’γ€–π’„π’π’”γ€—^𝟐⁑"ΞΈ" γ€— γ€—)) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (πŸ‘ 〖𝒄𝒐𝒔〗^πŸβ‘γ€–"ΞΈ" +γ€–πŸ 𝒄𝒐𝒔〗⁑〖"ΞΈ" βˆ’πŸγ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (πŸ‘ 〖𝒄𝒐𝒔〗^πŸβ‘γ€–"ΞΈ" +γ€–πŸ‘ π’„π’π’”γ€—β‘πœ½ γ€— βˆ’π’„π’π’”β‘πœ½ βˆ’πŸ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (3 γ€–π‘π‘œπ‘  〗⁑"ΞΈ" (cosβ‘πœƒ+1)βˆ’1(cosβ‘πœƒ+1)) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (3 π‘π‘œπ‘ β‘γ€–"ΞΈ" βˆ’1γ€— )(π‘π‘œπ‘ β‘γ€–"ΞΈ" +1γ€— ) Putting 𝒅𝑽/π’…πœ½ = 0 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (π‘π‘œπ‘ β‘γ€–"ΞΈ" +1γ€— ) (3 π‘π‘œπ‘ β‘γ€–"ΞΈ" βˆ’1γ€— ) = 0 𝐬𝐒𝐧 ΞΈ (cos ΞΈ + 1) (3 cos "ΞΈ" βˆ’ 1) = 0 𝐜𝐨𝐬 ΞΈ + 1 = 0 cos ΞΈ = βˆ’1 For cone, 0Β° < ΞΈ < 90Β° & cos ΞΈ is negative in 2nd & 3rd quadrant. So cos ΞΈ = βˆ’1 is not possible πŸ‘ 𝐜𝐨𝐬 ΞΈ βˆ’ 1 = 0 cos ΞΈ = 1/3 ΞΈ = cosβˆ’1 1/3 cos ΞΈ = 𝟏/πŸ‘ is possible So, 𝒄𝒐𝒔⁑"ΞΈ" = 𝟏/πŸ‘ Thus, H = r (1 + cos ΞΈ) H = r ("1 + " 1/3) H = πŸ’π’“/πŸ‘ Finding (𝒅^𝟐 𝑽)/(π’…πœ½^𝟐 ) 𝑑𝑉/π‘‘πœƒ = (1/3 πœ‹π‘Ÿ^3 sinβ‘πœƒ (3 cos2 πœƒ + 2 cosβ‘πœƒβˆ’ 1)) = 1/3 πœ‹π‘Ÿ^3 [cosβ‘πœƒ (3 cos^2β‘πœƒ+2 cosβ‘πœƒβˆ’1)+sinβ‘πœƒ (6 cosβ‘πœƒ (βˆ’π‘ π‘–π‘›)βˆ’2 sinβ‘πœƒ )] = 1/3 πœ‹π‘Ÿ^3 [3 cos^3β‘πœƒ+2 cos^2β‘πœƒβˆ’cosβ‘πœƒ+sinβ‘πœƒ (βˆ’6 cosβ‘πœƒ sinβ‘πœƒβˆ’2 sinβ‘πœƒ )] = 1/3 πœ‹π‘Ÿ^3 [3 cos^2β‘πœƒ+2 cos^2β‘πœƒβˆ’cosβ‘πœƒβˆ’6 sin^2β‘πœƒ cosβ‘πœƒβˆ’2 sin^2β‘πœƒ ] Now, 𝐜𝐨𝐬⁑𝜽=𝟏/πŸ‘ And sin2 "ΞΈ" = 1 βˆ’ cos2 "ΞΈ" = 1 βˆ’ (1/3)^2= 1 βˆ’ 1/9 = πŸ–/πŸ— Putting values in (𝒅^𝟐 𝑽)/(π’…πœ½^𝟐 ) (𝑑^2 𝑉)/(π‘‘πœƒ^2 ) = 1/3 πœ‹π‘Ÿ^3 [3(1/3)^3+2(1/3)^2βˆ’1/3βˆ’6(1/3)(8/9)βˆ’2(8/9)] = 1/3 πœ‹π‘Ÿ^3 [3(1/27)+2(1/9)βˆ’1/3βˆ’6(8/27)βˆ’2(8/9)] = 1/3 πœ‹π‘Ÿ^3 [1/9+ 2/9βˆ’1/3βˆ’16/9βˆ’16/9] = 1/3 πœ‹π‘Ÿ^3 [(βˆ’32)/9] = (βˆ’πŸ‘πŸπ…π’“^πŸ‘)/πŸπŸ• Thus, (𝑑^2 𝑉)/(π‘‘πœƒ^2 ) < 0 for 𝐜𝐨𝐬⁑𝜽=𝟏/πŸ‘ So, V is maximum for 𝐜𝐨𝐬⁑𝜽=𝟏/πŸ‘ Hence, altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is πŸ’π’“/πŸ‘ .

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.